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### Course: AP®︎/College Chemistry > Unit 15

Lesson 1: 2015 AP Chemistry free response questions- 2015 AP Chemistry free response 1a
- 2015 AP Chemistry free response 1b and c
- 2015 AP Chemistry free response 1d
- 2015 AP Chemistry free response 1e
- 2015 AP Chemistry free response 2a (part 1 of 2)
- 2015 AP Chemistry free response 2a (part 2/2) and b
- 2015 AP Chemistry free response 2c
- 2015 AP Chemistry free response 2d and e
- 2015 AP Chemistry free response 2f
- 2015 AP Chemistry free response 3a
- 2015 AP Chemistry free response 3b
- 2015 AP Chemistry free response 3c
- 2015 AP Chemistry free response 3d
- 2015 AP Chemistry free response 3e
- 2015 AP Chemistry free response 3f
- 2015 AP Chemistry free response 4
- 2015 AP Chemistry free response 5
- 2015 AP Chemistry free response 5a: Finding order of reaction
- 2015 AP Chemistry free response 6
- 2015 AP Chemistry free response 7

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# 2015 AP Chemistry free response 7

Energy necessary to recycle aluminum from aluminum oxide. From 2015 AP Chemistry free response 7.

## Want to join the conversation?

- In part (a) at1:33, how do you know if it is appropriate to use the 'molar quantity' rather than the molecular mass of Aluminium? (as in q = mass x specific heat capacity x temp change)? Is this only because we are given the MOLAR heat capacity?(7 votes)
- Taking a look at the units will help us here. The units of the molar hear capacity are (J/mol*K), so if we want to get Joules out the the equation, we are going to need to multiply by moles and Kelvin, so that out units cancel out. You're right in what you say- the "molar" heat capacity is a give away that we are going to have to work with numbers of moles here.(7 votes)

- Hi! A question about part B: Don't we have to take into account the phase and temperature difference between Al (s) (for extracting) and Al (l) (for purifying @ 933K) if we look at energy requirement?(4 votes)
- Actually you wouldn't worry about the difference of energy between the Al(extracted from Al2O3) and the Al(purified) because the purpose of the two processes is to get some aluminum (for making a cool sword of aluminum, for example), even though they're in different phases. What you could think about is how to get benefits from the already melted Aluminum to raise the temperature of other things (Although I don't know if and when it really would occurs). I hope this cleared out your doubts :) .(6 votes)

- Hi! So, The heat of fusion is the heat that is actually needed to both melt AND solidify?

Am I right? And also, for part a, why does the answer has to be in 2 sig.fig? Thanks.(3 votes)- Correct, the heat of fusion is the amount of heat required to melt a substance and the amount of heat given off by a substance solidifying.

The correct number of sigfigs is 2 because your final answer can't have more than the minimum number you started with. To see why this is true for 15,000 + 10,700, it may help to think about what 15,000 really means. Sigfigs are related to rounding — 15,000 means the real answer might be any amount from about 14,500 - 15,400. If we add 10,700 (really this is also a range) to each of those numbers we get 25,200 - 26,100. If we take the average of those we get 25,650. Rounded to 2 sigfigs that gives us 26,000, which is better than 25,650 because it conveys that we know the real answer is 20,000 + around 6,000 ...(5 votes)

- Why do we not multiply in the formula by Al in grams? Why do we multiply using moles instead?(3 votes)
- Sal used the heat equation to calculate the heat required to increase the temperature of one mol of Al from 298K to 933K. Since the specific heat capacity is in units of J/(mol)(K), he use this form of the heat equation: q = n(C)(deltaT) ; where n = number of moles. If the specific heat capacity had been given in J/gK, then he would have converted one mole to grams using the molar mass. And he would have used this form of the heat equation:

q = mass(C)(deltaT)(2 votes)

- dont you think that the heat of fusion of Al is the answer to part A?..because fusion means Solid to Liquid. So isn't 10.7KJ is the heat required to melt 1 mole of Al to evaporate impurities?...plz help(2 votes)
- Yes, but it is only a part of the answer. Just like you explained, we will eventually have to melt the Al (transform it from solid state to liquid state). But at what temperature does this physical transformation take place? It takes place at 933 K. The Al at our disposition is only at 298 K, so we need to reach the fusion temperature, which we will achieve by heating the Al. So the final answer for part A will have two parts: first, the heating of the Al from 298 K to 933 K; second, since we now have reached the necessary temperature for fusion, the actual fusion of Al.(3 votes)

- at5:06we divide by two, i thought we had to multiply. help please?(0 votes)
- The equation is Al₂O₃ → 2Al + ³/₂O₂; ΔH = 1675 kJ/mol of reaction.

The reaction involves**2**mol of Al.

∴ For 1 mol of Al,

Heat required = 1 mol Al × (1675 kJ/2 mol Al) = 837.5 kJ.(6 votes)

- hello , i can not see what formula do you use to calculate the needed heat to rise till 635K, and the other formula too , where we calculate the melting heat.

Thx.(3 votes)- Q= MCΔT Where

Q= Joules , M = Mass, C= Change in Temperature, Delta T= Change in temperature(0 votes)

- Why don't you just write the whole answer at2:37? Why round?(2 votes)
- Is molar heat capacity the same as specific heat?(2 votes)
- If he use Q= MCΔT, how come mols were used instead of mass? I expected him to use g/mol of Al from the periodic table.(0 votes)

## Video transcript

- [Voiceover] Aluminum
metal can be recycled from scrap metal by melting the metal to evaporate impurities. Calculate the amount of
head needed to purify 1.00 mole of aluminum originally
at 298 kelvin by melting it. The melting point of
aluminum is 933 kelvin. The molar heat capacity of aluminum is 24 joules per mole kelvin. And the heat of fusion of aluminum is 10.7 kilojoules per mole. Alright, so we didn't think
about the heat they'd need to raise its temperature from 298 kelvin we raise the temperature of
this one mole of aluminum from 298 kelvin to 933 kelvin, and we're going to do that by looking at the molar heat capacity. And then we're going to add to that the heat of fusion, the heat necessary to actually melt it. So let's do that in steps, so heat needed heat needed to raise temperature by how much are we raising the temperature by? We went from 298 to 933, so if we raise it by two we get 300 and then you have to
raise it by another 633. So that's going to be,
so we're raising our temperature by 635 kelvin. That's to get us to the melting point. So the heat needed to raise
temperature by 635 kelvin. Well, we have one mole one mole of aluminum I can write 1.00, so we can
have two significant digits, or three significant digits there. One mole of aluminum times the specific or the molar heat capacity I should say. So 24 joules per mole per mole kelvin times, so this right
over here would be the amount of heat to raise it by one kelvin but now lets multiply by, we
have to raise it by 635 kelvin 635 kelvin And the units work out,
moles cancel with moles. Kelvin cancels with
kelvin, and then we get to let's see let me get
my calculator out here. So 24 times 635 is equal to 15,240. and we're taking the
product of a bunch of stuff three significant figures,
two significant figures three significant figures,
so we should write only two significant figures
we'll round to 15,000 joules. So this is going to be 15,000 joules temperature to 933 kelvin. And then we have the heat of fusion heat to melt once at melting point. Melting point. Well that is going to be 1.00 mole 1.00 mole times the heat of fusion. times 10.7 kilojoules kilojoules per mole. So this is going to
give us 10.7 kilojoules or if we want to write it in joules, so this is 10.7 kilojoules which is equal to 10,700 joules. And so the total is going
to be the sum of these so we could say it is
25,700, but we only have two significant figures
here so we could round that so we could say 26,000 joules or I guess we could write 26 kilojoules. Alright, now part B. Part B. The equation for the overall
process of extracting aluminum from aluminum oxide
is shown below, alright? Which requires less energy,
recycling existing aluminum or extracting aluminum
from aluminum oxide? Justify your answer with a calculation. Alright, so this reaction right over here and this gives us the
heat necessary for a mole of the reaction, so if you
get a mole of aluminum oxide you put in this much heat, you're going to get 2.00 moles of aluminum and 1.5 moles of molecular oxygen. So the equation for the overall so to extract 1.00 mole 1.00 mole of aluminum from aluminum oxide requires so this much heat will produce 2.00 moles. So to do 1.00 mole you just
have to have half of that. Requires 1,675 kilojoules divided by two. So this is going to be
1,675 divided by two is equal to, let's see
it's going to be 837.5. 837.5 837.5 kilojoules, yep that's half here. and I had four significant digits here and I still have four significant digits. Extracting, doing it from, which requires less recycling existing aluminum or extracting it? Well to recycle 1.00 mole requires we just figured out, 26 kilojoules. 26 kilojoules. Everything here is in kilojoules, which is much less energy. Which is much less energy. And we are done.